To calculate the tension in the slack side of the belt (\(T_2\)), we will use the equations relating to belt tension in a crossed belt drive.
Given Data:
- Diameter of driving pulley (\(D_1\)) = 250 mm = 0.25 m
- Radius of driving pulley (\(R_1\)) = \(D_1 / 2 = 0.125\) m
- Speed of driving pulley (\(N_1\)) = 200 rpm
- Speed of driven pulley (\(N_2\)) = 250 rpm
- Power transmitted (\(P\)) = 30 kW = 30,000 W
- Coefficient of friction (\(\mu\)) = 0.25
- Angle of contact (\(\theta\)) = 192° = \(192 \times \frac{\pi}{180}\) rad = 3.356 rad
- Distance between pulley centers (\(L\)) = 800 mm = 0.8 m
Step 1: Calculate the belt tensions
In a crossed belt, the relationship between the tensions on the tight side (\(T_1\)) and the slack side (\(T_2\)) is given by:
\[ T_1 = T_2 e^{\mu \theta} \]
Step 2: Calculate the speed of the belt
The belt speed (\(v\)) can be calculated using the speed of the driving pulley:
\[ v = \frac{\pi D_1 N_1}{60} \]
Substituting the known values:
\[ v = \frac{\pi \times 0.25 \times 200}{60} \]
\[ v = \frac{50\pi}{60} \approx 2.618 , \text{m/s} \]
Step 3: Relate power to belt tension
Power transmitted can also be related to tension as follows:
\[ P = (T_1 - T_2) v \]
Rearranging gives:
\[ T_1 - T_2 = \frac{P}{v} \]
Substituting the known power and speed:
\[ T_1 - T_2 = \frac{30000}{2.618} \approx 11453.41 , \text{N} \]
Step 4: Solve for \(T_1\) and \(T_2\)
Now we have two equations:
- \(T_1 = T_2 e^{\mu \theta}\)
- \(T_1 - T_2 = 11453.41\)
Substituting equation 1 into equation 2:
\[ T_2 e^{\mu \theta} - T_2 = 11453.41 \]
Factor out \(T_2\):
\[ T_2 (e^{\mu \theta} - 1) = 11453.41 \]
Now calculate \(e^{\mu \theta}\):
\[ e^{\mu \theta} = e^{0.25 \times 3.356} \approx e^{0.839} \approx 2.316 \]
Substituting back in:
\[ T_2 (2.316 - 1) = 11453.41 \]
\[ T_2 (1.316) = 11453.41 \]
Solving for \(T_2\):
\[ T_2 = \frac{11453.41}{1.316} \approx 8692.23 , \text{N} \]
Conclusion
The tension in the slack side of the belt (\(T_2\)) is approximately 8692.23 N.
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